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This section is part of Discrete Mathematics. If you do not study that, you may not need to learn this.

An introductory example

A factory produces two types of drink: an energy drink and a refresher drink. The day’s output is to be planned. Each drink requires syrup, vitamin supplement and concentrated flavouring, as shown in the table. The last row in the table show how much of each ingredient is available for the day’s production.

Drink Type	Syrup (litres)	Vitamin Supplement (units)	Concentrated Flavouring (cc)
Energy Drink (5l)	3	2	30
Refresher Drink (5l)	1.25	1	20
Available	250	300	30

• Energy drinks sell for £1 per litre.
• Refresher drinks sell for £0.80 per litre.

• let x represent the litres of energy drink that can be produced.
• let y represent the litres of refresher drink that can be produced.

Syrup constraint:

Constraints

• x+y\le1000
• 2x+y\le1500
• 6x+4y\le4800
• x\ge0
• y\ge0

Objective function

• x+0.8y.

We have now formulated the linear programming problem.

Solving

• A: x=0, y=1000: 1(0)+0.8(1000)=800
• B: x=400, y=600: 1(400)+0.8(600)=880
• C: x=600, y=300: 1(600)+0.8(300)=840
• D: x=750, y=0: 1(750)+0.8(0)=750

The maximum profit is £880 at point B, where 400 litres of energy drink and 600 litres of refresher drink are produced.

Another example

A factory produces two toys: a bicycle and a truck. Each toy requires time on a lathe and time on an assembler, as shown in the table. The last row in the table shows how much time is available on each machine for the day’s production.

Toy Type	Lathe (hours)	Assembler (hours)
Bicycle	2	2
Truck	1	3
Available	16	12

Constraints

• 2x+y\le16 (lathe)
• 2x+3y\le24 (assembler)
• x\ge0
• y\ge0

Objective function

• P=16x+14y

We have now formulated the linear programming problem.

A third example

A wall is to be built using two types of bricks of equal thickness. The wall is to be 3 metres high, 100 metres long and the thickness of a single brick. The large type of brick is of height 10cm and length 20cm. The small type of brick is of height 5cm and length 10cm.

Ignoring the mortar between the bricks, the constraint 4L+s\ge60000 ensures that there are enough bricks to build the wall. Identify the meanings of the variables L and S, and explain why the constraint works.

• L represents the number of large bricks used.
• S represents the number of small bricks used.
• Each large brick has a face area of 10cm\times20cm=200cm^2, or 0.02m^2.
• Each small brick has a face area of 5cm\times10cm=50cm, or 0.005m^2.
• So the large bricks are 4 times the area of the small bricks.
• The total area of the wall is 3m\times100m=300m^2.
• This means that the total area of the bricks must be at least 300m^2.
• Therefore, the constraint 0.02L+0.005S\ge300.
• This constraint is equivalent to 4L+S\ge60000 when both sides are multiplied by 200.

Each large brick needs 0.5 litres of mortar. Each small brick requires 0.25 litres of mortar.

Given that 10000 litres of mortar are available, write down a constraint for the number of large and small bricks that can be used.

• 0.5L+0.25S\le10000.
• Simplifies to 2L+S\le40000

Large bricks cost 25p each to produce, and small bricks cost 5p each to produce. Formulate a linear programming to find how many bricks of each type should be used to build the wall as cheaply as possible.

• Let L represent the number of large bricks used.
• Let S represent the number of small bricks used.
• Constraints:
  • 4L+S\ge60000
  • 2L+S\le40000
  • L\ge0
  • S\ge0
• Objective function:
  • C=0.25L+0.05S

We have now formulated the linear programming problem.

Use a graphical method to solve this. Interpret your solution.

• A: x=10000, y=20000: 0.25(10000)+0.05(20000)=2500+1000=£3500
• B: x=15000, y=0: 0.25(15000)+0.05(0)=3750+0=£3750
• C: x=20000, y=0: 0.25(20000)+0.05(0)=5000+0=£5000

The cheapest way to build the wall is to use 10000 large bricks and 20000 small bricks, which will cost £3500.

A fourth example

A landscaping project is being planned to cover an area of 2000m^2. £3700 has been allocated to buy plants for the project. Trees cost £30 each and require 30m^2 of space. Shrubs cost £9 each and require 4m^2 of space. It has been decided that at least 75 shrubs should be planted. Trees are five times as beneficial to the environment as shrubs. How many trees and shrubs should be planted to maximise the environmental benefit?

• Let x represent the number of trees planted.
• Let y represent the number of shrubs planted.

Objective function

• Maximise 5x+y.

Constraints

• x\ge0 (can’t have negative trees)
• y\ge75 (minimum 75 shrubs)
• 30x+4y\le2000 (space)
• 30x+9y\le3700 (budget)

We have now formulated the linear programming problem.

Solving

• A: x=0, y=411.111... (but we can only have whole shrubs, so y=411):
  • 5(0)+411=411
• B: x=21.333..., y=340 (but we can only have whole trees, so x=21):
  • 5(21)+340=105+340=445
• C: x=56.666..., y=75 (but we can only have whole trees, so x=56):
  • 5(56)+75=280+75=355

The maximum environmental benefit is 445 (arbitrary units) at point B. This involves planting 21 trees and 340 shrubs.

A more optimal solution?

Because we rounded down the number of trees, it is now possible that we could fit some more shrubs into the budget and space available. We have to check this to ensure that our solution is optimal.

• With 21 trees, we have used £630 of the budget and 630m^2 of space.
• This leaves £3070 and 1370m^2 for shrubs.
• Each shrub costs £9 and requires 4m^2 of space.
• The budget allows for a maximum of \frac{3070}{9}=341.111... shrubs.
• The space allows for a maximum of \frac{1370}{4}=342.5 shrubs.
• Therefore, we can afford to plant 341 shrubs.

Aha! We can plant another shrub, increasing our environmental benefit to 5(21)+341=105+341=446 (arbitrary units).

You need to consider this in linear programming problems where the values are not integers. In exams, it is unlikely that you will be given a problem where this can happen, but it is worth being aware of.